hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams among the TAs if all that matters is how many exams go to each TA? (b) Now suppose it matters which students' exams go to which TAs. How many ways are there to distribute the exams? (c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Question

Jarrett

Mathematics

30 May, 19:55

hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams among the TAs if all that matters is how many exams go to each TA? (b) Now suppose it matters which students' exams go to which TAs. How many ways are there to distribute the exams? (c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

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Answers (1)

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Lillian Little · Answer 1

a. 205320

b. 34220

c. 60! / (35) ! (25) ! + 60! / (40) ! (20) ! + 60! / (45) ! (15) !

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r) !

n = number of things to choose from

r = Choosing r number

60P3 = 60! / (60 - 3) !

(60) (59) (58) (57) ! / (57) !

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! / (n - r) ! r!

60C3 = 60! / (60 - 3) ! 3!

= 60! / 57! 3!

= 60 * 59 * 58 / 3 * 2 * 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35) ! (25) ! + 60! / (40) ! (20) ! + 60! / (45) ! (15) !